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Sergeeva-Olga [200]

4 years ago

7

As 8 hundreds is 8 thousands

1 answer:

Mice21 [21]4 years ago

6 0

No 8 hundreds is not the same as 8 thousands unless you use 80 hundreds

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What is the answer lol somebody help me please and thank you !

PilotLPTM [1.2K]

I'm guessing that you know about 45-45-90 and 30-60-90 triangles.

In 45-45-90 triangles, the legs are in a ratio such that the hypotenuse is $\sqrt{2}$ times the legs.

In 30-60-90 triangles, the legs are in a ratio of 1: $\sqrt{3}$ :2

In the picture given we have been given a value of the side of the 45-45-90 triangle. We can deduce that the value of the hypotenuse is 9 $\sqrt{2}$ because of the ratio of the sides of a 45-45-90 triangle.

The 30-60-90 triangle shares the same hypotenuse, and so we now know a value of one of its sides as well. The ratio of the side of value 'x' and the hypotenuse is $\sqrt{3}$ : 2

We can create an equation to solve for x using these ratios:

$\frac{\sqrt{3} }{2}$ = $\frac{x}{9\sqrt{2} }$

Cross multiply:

2x = 9 $\sqrt{6}$

x = $\frac{9\sqrt{6} }{2}$

Option C

4 0

3 years ago

Building a has 7,500 ft of an office space for 320 employees building b has 9,500 square feet of office space for 370 employees

iren2701 [21]

Answer:

I believe the correct answer is Building B

Step-by-step explanation:

Building A is 23.45 per employee and Building B is 25.67 per employee

4 0

3 years ago

What is <br>(2x3) x6+17-4=

iren [92.7K]

6x6+17-4
36+17-4
52-4
48

4 0

3 years ago

Read 2 more answers

Find the least common multiple of x3 - x2 + x - 1 and x2 - 1. Write the answer in factored form. A. (x + 1)2(x - 1) B. (x + 1)(x

WINSTONCH [101]

Hello,
Answer B

$x^3-x^2+x-1=(x-1)(x^2+1)\\
x^2-1=(x+1)(x-1)\\
lcm=(x-1)(x+1)(x^2+1)\\
$

8 0

4 years ago

Read 2 more answers

The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the

vitfil [10]

Answer:

The probability is 0.4207

Step-by-step explanation:

The probability of a home-based computer having access to on-line services is p = 0.2 (data from the exercise)

Then, the probability of a home-based computer not having access to on-line services is p = 1 - 0.2 = 0.8

We are going to use this probability (p = 0.8) to solve the exercise.

Let's define the random variable X

X : ''Number of home-based computers not having access to on-line services''

X can be modeled as a binomial random variable

X ~ Bi(p,n)

X ~Bi(0.8,25)

Where p is the success probability and n is the number of Bernoulli independent experiments we are taking place.

We are going to count ''a success'' as a computer not having access to on-line services.

The binomial probability function is :

$P(X=x)=(nCx)p^{x}(1-p)^{n-x}$

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as

$nCx=\frac{n!}{x!(n-x)!}$

p is the success probability and n the number of Bernoulli independent experiments taking place.

In our exercise,

$p=0.8\\n=25$

We are looking for :

$P(X>20)=P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)$

$P(X>20)=(25C21)0.8^{21}0.2^{4}+(25C22)0.8^{22}0.2^{3}+(25C23)0.8^{23}0.2^{2}+(25C24)0.8^{24}0.2^{1}+(25C25)0.8^{25}0.2^{0}$

$P(X>20)=0.1867+0.1358+0.0708+0.0236+0.8^{25}$

$P(X>20)=0.4207$

Finally, the probability of finding that more than 20 of 25 home-based computers do not have access to on-line services is 0.4207

6 0

3 years ago